Question

A 0.248 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.148 kg puck moving initially along the x axis with a speed of 2.82 m/s. After the collision, the 0.148 kg puck has a speed of 1.82 m/s at an angle of 25◦ to the positive x axis. Find the fraction of kinetic energy lost in the collision.

Answer 1

Answer:

The fraction of kinetic energy lost in the collision is 0.25 .

Explanation:

We know , momentum will be conserved along x direction .

Therefore ,

$0.248 \times 0+0.148\times 2.82 = 0.148\times 1.82 \times sin \ 25^{\circ} + 0.248\times v\\\\0.42=0.11+0.248 \times v\\\\v=1.25\ m/s$

Now , fraction lost in kinetic energy is :

$Loss =\dfrac{K.E_i-K.E_f}{K.E_i}\\\\Loss=1-\dfrac{K.E_f}{K.E_i}\\\\Loss=1-\dfrac{\dfrac{0.148\times1.82^2}{2}+\dfrac{0.248 \times 1.25^2}{2}}{\dfrac{0.148\times 2.82^2}{2}+\dfrac{0.248\times 0^2}{2}}\\\\Loss =0.25$

Therefore , the fraction of kinetic energy lost in the collision is 0.25 .